Problem: Given a string of length n s = s1, s2,…, sn and m queries. Each query qk (1 ≤ k ≤ m) is one of four types, "L ++", "L-", "R ++", "R-", and l [for the kth query qk. k] and r [k] are defined below.

* L ++: l [k] = l [k-1] + 1, r [k] = r [k-1]

* L-: l [k] = l [k-1] -1, r [k] = r [k-1]

* R ++: l [k] = l [k-1], r [k] = r [k-1] +1

* R-: l [k] = l [k-1], r [k] = r [k-1] -1




However, l [0] = r [0] = 1.

At this time, how many kinds of character strings are created for m substrings sl [k], sl [k] + 1,…, sr [k] -1, sr [k] (1 ≤ k ≤ m). Answer if you can.

Constraints

* The string s consists of a lowercase alphabet

* 1 ≤ n ≤ 3 x 105

* 1 ≤ m ≤ 3 × 105

* qk (1 ≤ k ≤ m) is one of "L ++", "L-", "R ++", "R-"

* 1 ≤ l [k] ≤ r [k] ≤ n (1 ≤ k ≤ m)

Input

Input is given in the following format

> n m
> s
> q1
> q2
>…
> qm
>

Output

Output the solution of the problem on one line

Examples

Input

5 4
abcde
R++
R++
L++
L--


Output

3


Input

4 6
abab
R++
L++
R++
L++
R++
L++


Output

4


Input

10 13
aacacbabac
R++
R++
L++
R++
R++
L++
L++
R++
R++
L--
L--
R--
R--


Output

11

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
#define rep(i,a,b) for(int i=a;i<b;i++)

typedef long long ll;
struct RollingHash {
	static const ll mo0 = 1000000007, mo1 = 1000000009;
	static ll mul0, mul1;
	static const ll add0 = 1000010007, add1 = 1003333331;
	static vector<ll> pmo[2];
	vector<int> s; int l; vector<ll> hash_[2];
	void init(vector<int> s) {
		this->s = s; l = s.size(); int i, j;
		hash_[0] = hash_[1] = vector<ll>(1, 0);
		if (!mul0) mul0 = 10009 + (((ll)&mul0) >> 5) % 259, mul1 = 10007 + (((ll)&mul1) >> 5) % 257;
		if (pmo[0].empty()) pmo[0].push_back(1), pmo[1].push_back(1);
		for (int i = 0; i < l; i++) hash_[0].push_back((hash_[0].back()*mul0 + add0 + s[i]) % mo0);
		for (int i = 0; i < l; i++) hash_[1].push_back((hash_[1].back()*mul1 + add1 + s[i]) % mo1);
	}
	pair<ll, ll> hash(int l, int r) { // s[l..r]
		if (l>r) return make_pair(0, 0);
		while (pmo[0].size()<r + 2)
			pmo[0].push_back(pmo[0].back()*mul0%mo0), pmo[1].push_back(pmo[1].back()*mul1%mo1);
		return make_pair((hash_[0][r + 1] + (mo0 - hash_[0][l] * pmo[0][r + 1 - l] % mo0)) % mo0,
			(hash_[1][r + 1] + (mo1 - hash_[1][l] * pmo[1][r + 1 - l] % mo1)) % mo1);
	}
};
vector<ll> RollingHash::pmo[2]; ll RollingHash::mul0, RollingHash::mul1;

int N, M;

int main()
{
	cin.tie(0);
	ios::sync_with_stdio(false);

	while (cin >> N >> M)
	{
		string s; cin >> s;

		RollingHash rh;
		vector<int> v(s.length());
		rep(i, 0, s.length()) v[i] = s[i] - 'a';
		rh.init(v);

		int l = 0, r = 0;
		vector<pair<ll, ll> > vp;
		rep(i, 0, M)
		{
			string q; cin >> q;
			if (q == "L++")
				l++;
			else if (q == "L--")
				l--;
			else if (q == "R++")
				r++;
			else
				r--;

			vp.push_back(rh.hash(l,r));
		}

		sort(vp.begin(), vp.end());

		int ans = 1;
		pair<ll, ll> back = vp[0];

		rep(i, 1, M)
		{
			if (back != vp[i]) ans++;
			back = vp[i];
		}

		cout << ans << endl;
	}
}